# Show that an=f(1)f(2).........f(n) convergent, n > or equal to 1f(x)=2x+5/3x+4

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You should use the ratio criterion to test the convergence of the series such that:

`lim_(n-gtoo) (a_(n+1))/(a_n)`

Since you know `a_n` , you need to form the series `a_(n+1)` based on a_n definition such that:

`lim_(n-gtoo) (f(1)*f(2)*....*f(n)*f(n+1))/(f(1)*f(2)*....*f(n))`

Reducing like terms yields:

`lim_(n-gtoo) f(n+1)`

Plugging x = n+1 in equation of function f(x) yields:

`f(n+1) = (2(n+1)+5)/(3(n+1)+4)=gtf(n+1) = (2n+7)/(3n+7)`

Evaluating the limit yields:

`lim_(n-gtoo) f(n+1) = lim_(n-gtoo) (2n+7)/(3n+7)`

Notice that both numerator and denominator have like order hence the limit of the fraction is the ratio of leading coefficients such that:

`lim_(n-gtoo) (2n+7)/(3n+7) = 2/3`

**Since the ratio `2/3 lt 1` , hence the series `a_n ` converges absolutely.**